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\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{实变函数练习 4.2 - 4.4 \\ 叶戈罗夫定理、可测函数的构造、依测度收敛 }
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\date{2024 年 5 月 6 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 01 叶戈罗夫定理
设可测集 $E$ 的测度有限。设 $\{f_n(x)\}$ 是 $E$ 上的一列可测函数，且在 $E$ 上几乎处处收敛于有限函数 $f(x)$. 则对任意 $\delta>0$, 存在子集 $E_\delta\subseteq E$, 使得 $m(E-E_\delta)<\delta$, 且 $\{f_n(x)\}$ 在 $E_\delta$ 上是一致收敛的。

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\item  %Problem 02 卢津定理1
设 $f(x)$ 是可测集 $E$ 上几乎处处有限的可测函数，则对任意 $\delta>0$, 存在闭子集 $F_\delta\subseteq E$, 使得 $f(x)$ 在 $F_\delta$ 上是连续函数，且 $m(E-F_\delta)<\delta$.

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\item  %Problem 03 卢津定理2
设 $f(x)$ 是可测集 $E\subseteq \mathbb{R}$ 上几乎处处有限的可测函数，则对任意 $\delta>0$, 存在闭子集 $F\subseteq E$ 以及整个 $\mathbb{R}$ 上的连续函数 $g(x)$, 使得在 $F$ 上 $g(x)=f(x)$, 且 $m(E-F)<\delta$. 
%此外还可要求 $\sup_\mathbb{R}\{g(x)\} = \sup_\mathbb{R}\{f(x)\}$ 以及 $\inf\mathbb{R}\{g(x)\} = \inf\mathbb{R}\{f(x)\}$.

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\item  %Problem 04
设 $\{f_n(x)\}$ 是可测集 $E$ 上的一列几乎处处有限的可测函数。
设 $f(x)$ 是 $E$ 上几乎处处有限的可测函数。
什么时候称 $\{f_n(x)\}$ 依测度收敛于 $f(x)$ ？

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\item  %Problem 05 例子4.1
举例说明函数列可以处处不收敛，但是依测度收敛。也可以处处收敛，但不是依测度收敛。

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\item  %Problem 06 定理4.1：里斯定理
设可测集 $E$ 上的函数列 $\{f_n(x)\}$ 依测度收敛于 $f(x)$. 证明存在子列 $f_{n_i}(x)$ 几乎处处收敛于 $f(x)$.

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\item  %Problem 07  定理4.2： 勒贝格定理
设可测集 $E$ 的测度有限。设 $\{f_n(x)\}$ 是 $E$ 上的几乎处处有限的可测函数列，$f(x)$ 是 $E$ 上的几乎处处有限的函数，且 $\{f_n(x)\}$ 在 $E$ 上几乎处处收敛于 $f(x)$. 则 $f_n(x)$ 依测度收敛于 $f(x)$.

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\item  %Problem 08 定理4.3
设可测集 $E$ 上的函数列 $\{f_n(x)\}$ 依测度收敛于 $f(x)$, 也依测度收敛于 $g(x)$. 证明 $f(x)=g(x)$ 在 $E$ 上几乎处处成立。

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\item  %Problem 09 例子4.3
设 $f(x)$ 是定义在 $E\subseteq\mathbb{R}$ 上的几乎处处有限的可测函数。证明存在定义在 $\mathbb{R}$ 上的一列连续函数 $\{g_n(x)\}$, 使得 $\lim\limits_{n\to\infty} g_n(x)=f(x)$ 在 $E$ 上几乎处处成立。

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\item  %Problem 10 习题9
设在可测集 $E$ 上，$\{f_n(x)\}$ 依测度收敛于 $f(x)$, 且对任意正整数 $n$, $f_n(x)\le g(x)$ 在 $E$ 上几乎处处成立。证明 $f(x)\le g(x)$ 在 $E$ 上几乎处处成立。

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\item  %Problem 11 习题13
设 $m(E)<\infty$. 设 $\{f_n(x)\}$ 与 $\{g_n(x)\}$ 是 $E$ 上的两个有限的可测函数列，分别依测度收敛于 $f(x)$ 与 $g(x)$. 证明 $\{f_n(x)g_n(x)\}$ 依测度收敛于 $f(x)g(x)$.

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\item  %Problem 12 习题16
设 $m(E)<\infty$, 设 $\{f_n(x)\}$ 是 $E$ 上有限的可测函数列。
记 $$g_n(x) = \sup\{ |f_k(x)|: k\ge n\},$$
证明 $\lim\limits_{n\to\infty} f_n(x)=0$ 在 $E$ 上几乎处处成立的充分必要条件是 $\{g_n(x)\}$ 依测度收敛于零。

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\end{enumerate}


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\end{document}

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